SME
Numerical Methods for Engineers
Root finding
Use this for “Take one Newton step for x³ − x − 3 = 0 from x₀ = 2, and find the approximate error.”
Chart
Newton: each tangent meets the axis at the next x: x₀ = 2, x₁ = 1.727
This is an example. Type your own values to start a new problem.
f(x)
=
c₃
x³
+
c₂
x²
+
(
c₁
)
x
+
(
c₀
)
x₁
=
x₀
−
f(
)
f′(
)
=
f(x₀)
Function value at x₀
worked out (not typed)
f′(x₀)
Slope at x₀
worked out (not typed)
εₐ
Approximate error
calculated · %
Formulas
f
(
x₀
)
=
c₃
×
x₀
3
+
c₂
×
x₀
2
+
c₁
×
x₀
+
c₀
3
=
1
×
2
3
+
0
×
2
2
+
(−1)
×
2
+
(−3)
f
′(
x₀
)
=
3
×
c₃
×
x₀
2
+
2
×
c₂
×
x₀
+
c₁
11
=
3
×
1
×
2
2
+
2
×
0
×
2
+
(−1)
x₁
=
x₀
−
f
(
x₀
)
f
′(
x₀
)
1.7273
=
2
−
3
11
εₐ
=
|
x₁
−
x₀
|
÷
|
x₁
|
×
100
15.7895 = |1.7273 − 2| ÷ |1.7273| × 100
Assumptions
•
f(x) = c₃x³ + c₂x² + c₁x + c₀; its slope f′(x) comes from the power rule.
•
One step follows the tangent at x₀ down to the axis: x₁ = x₀ − f(x₀) ÷ f′(x₀).
•
The approximate error εₐ = |x₁ − x₀| ÷ |x₁| says how much the step changed x.
•
Further steps repeat the same rule from x₁; this page works the first one.
Step-by-step
WE KNOW
c₃ = 1, c₂ = 0, c₁ = −1, c₀ = −3, x₀ = 2
FIND
function value at x₀ (f(x₀)), slope at x₀ (f′(x₀)), next guess (x₁), approximate error (εₐ)
Step 1 · Find function value at x₀ (f(x₀))
Use f(x₀) = c₃ × x₀³ + c₂ × x₀² + c₁ × x₀ + c₀
Put the starting guess into f.
f
(
x₀
)
=
1
×
2
3
+
0
×
2
2
+
(−1)
×
2
+
(−3)
f
(
x₀
)
=
1
×
8
+
0
×
4
−
1
×
2
−
3
f
(
x₀
)
=
8
−
2
−
3
f
(
x₀
)
=
3
Step 2 · Find slope at x₀ (f′(x₀))
Use f′(x₀) = 3 × c₃ × x₀² + 2 × c₂ × x₀ + c₁
Power
rule
on
each
term:
bring
the
power
down
in
front
and
lower
it
by
1;
c₀
has
slope
0.
f
′(
x₀
)
=
3
×
1
×
2
2
+
2
×
0
×
2
+
(−1)
f
′(
x₀
)
=
3
×
4
+
0
×
2
−
1
f
′(
x₀
)
=
12
−
1
f
′(
x₀
)
=
11
Step 3 · Find next guess (x₁)
Use x₁ = x₀ − f(x₀) ÷ f′(x₀)
The
tangent
at
x₀
has
slope
f
′(
x₀
);
it
meets
the
axis
f
(
x₀
)
f
′(
x₀
)
to
the
side
of
x₀
.
x₁
=
2
−
3
11
x₁
=
2
−
0.2727
x₁
=
1.727273
Step 4 · Find approximate error (εₐ)
Use εₐ = |x₁ − x₀| ÷ |x₁| × 100
The size of the step as a percent of the new guess; iterate until it is under your tolerance.
εₐ = |1.727273 − 2| ÷ |1.727273| × 100
εₐ = |−0.2727| ÷ |1.7273| × 100
εₐ
=
0.2727
1.727
×
100
εₐ = 0.1579 × 100
εₐ = 15.7895%
Check
Put the numbers back in. Both sides should match.
3
=
1
×
2
3
+
0
×
2
2
+
(−1)
×
2
+
(−3)
✓
11
=
3
×
1
×
2
2
+
2
×
0
×
2
+
(−1)
✓
1.727273
=
2
−
3
11
✓
15.7895 = |1.727273 − 2| ÷ |1.727273| × 100 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Bisection: halve the bracket
Use this for “Bisect x³ − x − 3 = 0 on [1, 2] once, then say how many halvings reach a tolerance of 0.0001.”
Secant method: one step
Use this for “Take one secant step for x³ − x − 3 = 0 from x₀ = 1.5 and x₁ = 2.”
Bracketing or open methods
Use this for “Is the secant method a bracketing method or an open one?”